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Thread: prove there exists an inverse function

  1. #1
    Junior Member
    Nov 2012

    prove there exists an inverse function

    Hello. I'm quite new to calculus and frankly speaking I'm an autodidact. Could you tell me how to solve this:

    $\displaystyle f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{sinx} \in R$

    Prove that there exists a inverese of $\displaystyle f$. In what set is $\displaystyle f^{-1}$ differentiable? Calculate $\displaystyle f^{-1'}$

    I would really appreciate a thorough explanation. Thank you.
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  2. #2
    Senior Member jakncoke's Avatar
    May 2010

    Re: prove there exists an inverse function

    Well, you could easily show that on the interval $\displaystyle [\frac{\pi}{2}, \pi) $, f is monotonously increasing, since $\displaystyle f'(x) $ exists on the interval, it is continuous on the interval. A continous function which is monotously increasing or decreasing is one to one on that interval and thus also has an inverse. Now if f is differentiable on the interval, $\displaystyle f^{-1} $ is also differentiable given that $\displaystyle f'(x) \not = 0 $ at any point on the interval. You can verify that this is true. Thus $\displaystyle f^{-1} $ is differentiable over $\displaystyle [\frac{\pi}{2}, \pi) $
    Last edited by jakncoke; Dec 18th 2012 at 10:31 AM.
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